Answer
a. $\frac{p}{q}=\pm1,\pm\frac{1}{2}$
b. positive real zeros: $2,0$. negative real zeros: 1.
c. $x=\frac{1}{2}$
d. $x=\frac{-5\pm\sqrt {29}}{2}$ (and the zero from c).
Work Step by Step
a. Given the function
$f(x)=2x^3+9x^2-7x+1$
we have $p=\pm1, q=\pm1,\pm2$, and the possible rational zeros are $\frac{p}{q}=\pm1,\pm\frac{1}{2}$
b. There are 2 sign changes in the equation and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be $2,0$
We examine
$f(-x)=2(-x)^3+9(-x)^2-7(-x)+1=-2x^3+9x^2+7x+1$
We can find 1 sign change and based on Descartes’s Rule of Signs, we can determine that there is 1 negative real zero.
c. We can use synthetic division to find a zero as $x=\frac{1}{2}$, as shown in the figure.
d. Part (c) gives a quotient $2x^2+10x-2=2(x^2+5x-1)$; thus the other two zeros (roots) are $x=\frac{-5\pm\sqrt {29}}{2}$.
